Question

In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 75, when a α-particle of 5 MeV energy impinges on it before it comes momentarily to rest and reverses its direction.

How will the distance of closest approach be affected when the kinetic energy of the α-particle is doubles?

Answer 1

Let r₀ be the centre to centre distance between the alpha-particle and nucleus when the α-particle is at its stopping point.

Now`,E_k = 1/(4piepsi_0) ((2e)(ze))/r_0 or r_0 =1/(4piepsi_0 )((2e)(ze))/E_k`

given E_k = 5 × 10⁶eV

= 5 × 10⁶× 1.6 × 10^-19V

Z = 75

`r_0 = (9 xx 10^9 xx 75 xx 2(1.6 xx 10^-19)^2)/(5 xx 10^6 xx 1.6 xx 10^-19)`

   `  =(3456 xx 10 xx^9 xx 10^-38)/(8 xx 10^-13)`

    `= 432 xx 10^(+22 -38)`

   `=432 xx 10^-16 m`

  ` =43.2 xx 10^-15 m =43.2 fm`

Since distance of closet approach (r0) is given as

`r_0 = 1/(4piepsi_0) ((2e)(ze))/E_k`

Where Ek is kinetic energy r0 is inversely proportional to Ek.

So when kinetic energy of the α-particle is doubled the distance between closet approach r0 is halved.